Consider the following cumulative distribution function $F(x)$ for $x \in [0, 2]$
Figure is:
a. what's density function $f(x)$ corresponding to this $F(x)$.
b. whether a random variable with this distribution has positive or negative expectation? why?
c. Compute the probability of the event: $A = \lbrace\omega: X(\omega) \in (0.6, 0.65]\rbrace$.
continued for part c.
From my understanding, probability density function $f(x)$ is the derivative of $F(x)$.
So,
(a) $f(x) = \begin{cases} 1 &\text{when}&\quad 0 ≤ x < 0.5; \\ 0&\text{when}&0.5 ≤ x < 1.5; \\1&\text{when}&1.5 ≤ x < 2; \end{cases}$
since $F(x)$ for $x \in [0, 2]$, so we only conside $[0, 2]$ for $f(x)$ too.
(b) Since $\Bbb E(x) = \sum_x f(x)\cdot p(x)$, $f(x)\geq 0, 0\leq p(x)\leq 1$, Hence, I guess Expectation must be positive?
(c) I've just lost here. How could we calculate the specific event of $X(\omega) \in (0.6, 0.65]$, when we have PDF or CDF?
Thanks a lot!
(a) Almost. The pdf equals one over those two intervals, and zero anywhere else. (ie: Not just between them.) $$f(x) = \begin{cases}1 &:& (0\leq x < 0.5) \vee(1.5\leq x< 2)\\ 0 &:& \text{otherwise} \end{cases}$$
This is a uniform continuous distribution over the interval $[0;0.5)\cup[1.5;2)$
(b) That is not the formula for expectation of a continuous random variable with density function $f(x)$. Regardless, you should be able to establish the expectation of the random variable by inspecting the graph.
(c) The event $\lbrace \omega\in\Omega : X(\omega)\in(0.6;0.65]\rbrace$ is also know as the event of: $0.6<X\leq 0.65$. Either by inspecting the graph, or from the probability density function you should be able to determine the probability that the random variable may be within this interval.
(d) A cumulative distribution function has certain properties. Does this graph satisfy all of them? List them and check.